Question

If the projection of \[ \vec a = 5\hat i+\hat j+\lambda\hat k \] on \[ \vec b = 2\hat i+6\hat j+3\hat k \] is \(4\) units, then \(\lambda=\)

• A. \(6\)
• B. \(4\)
• C. \(5\)
• D. \(3\)

Hint:

Scalar projection formula: \[ \text{proj}_{\vec b}\vec a = \frac{\vec a\cdot\vec b}{|\vec b|} \] Always compute the dot product first and then divide by the magnitude.

Answer 1

The Correct Option is B

Concept: The scalar projection of vector \(\vec a\) on vector \(\vec b\) is given by: \[ \text{Projection of }\vec a\text{ on }\vec b = \frac{\vec a\cdot\vec b}{|\vec b|} \] where: \[ \vec a\cdot\vec b = a_1b_1+a_2b_2+a_3b_3 \] and \[ |\vec b| = \sqrt{b_1^2+b_2^2+b_3^2} \]

Step 1: Compute the dot product. Given: \[ \vec a=(5,1,\lambda) \] and \[ \vec b=(2,6,3) \] Thus, \[ \vec a\cdot\vec b = 5(2)+1(6)+\lambda(3) \] \[ = 10+6+3\lambda \] \[ = 16+3\lambda \]

Step 2: Find the magnitude of \(\vec b\). \[ |\vec b| = \sqrt{2^2+6^2+3^2} \] \[ = \sqrt{4+36+9} \] \[ = \sqrt{49} \] \[ =7 \]

Step 3: Use the projection formula. Given projection \(=4\), \[ \frac{16+3\lambda}{7}=4 \] Multiply both sides by \(7\): \[ 16+3\lambda=28 \] \[ 3\lambda=12 \] \[ \lambda=4 \] Hence, \[ \boxed{4} \] Therefore the correct answer is: \[ \boxed{(B)\ 4} \]